11 Chi-Square Tests

Categorical Data

Chi-Square Tests

Goodness-of-Fit Tests

Tests of Independence

Relative Risk

Objectives

Upon successful completion of this lesson, you should be able to:

Construct a chi-square probability distribution plot in StatKey or Minitab.
Determine when a chi-square goodness-of-fit test and chi-square test of independence should be conducted.
Compute and interpret expected counts.
Conduct chi-square tests by hand and using Minitab.
Calculate and interpret relative risk.

This lesson corresponds to Chapter 7 in the Lock⁵ textbook.

In this lesson we will learn how to compare the proportions of more than two independent groups and how to test for a relationship between two categorical variables. In this lesson you will also review risk and odds and learn about relative risk.

Before we begin, let’s review the various classifications of variables from Lesson 1.

Definition 11.1 (Categorical) Names or labels (i.e., categories) with no logical order or with a logical order but inconsistent differences between groups. Also known as qualitative.

Definition 11.2 (Quantitative) Numerical values with magnitudes that can be placed in a meaningful order with consistent intervals. Also known as numerical.

In this lesson, we will be examining methods for analyzing categorical variables.

11.1 Review

In this lesson you may need to use Minitab to construct frequency tables or two-way contingency tables. We’ll start by reviewing these procedures. You will need to construct probability distribution plots for chi-square distributions. In earlier lessons we constructed probability distribution plots for z-, t-, and F-distributions; the procedure is similar for a chi-square distribution. We will also review conditional probabilities and the term independence in this section.

Frequency Tables

The following example was first presented in Lesson 2.

It uses following data set (from College Board): (sat_data.mpx)

Minitab: Frequency Tables

To create a frequency table in Minitab:

Open the Minitab file: (sat_data.mpx)
Select Stat > Tables > Tally Individual Variables.
Insert the variable Region into the Variable box on the right.
Under Statistics, check Counts and Percents.
Choose OK.

This should result in the following frequency table:

Tally

Region	Count	Percent
ENC	5	9.80
ESC	4	7.84
MA	3	5.88
MTN	8	15.69
NE	6	11.76
PAC	5	9.80
SA	9	17.65
WNC	7	13.73
WSC	4	7.84
N=	51

Two-Way Contingency Table

Recall from Section 2.1.2 that a two-way contingency table is a display of counts for two categorical variables in which the rows represented one variable and the columns represent a second variable. The starting point for analyzing the relationship between two categorical variables is to create a two-way contingency table. When one variable is obviously the explanatory variable, the convention is to use the explanatory variable to define the rows and the response variable to define the columns; this is not a hard and fast rule though.

Minitab: Constructing a Two-Way Contingency Table

Open the data set: (class_survey.mpx)
Select Stat > Tables > Cross Tabulation and Chi-square.
Select Raw data (categorical variable) from the dropdown menu.
Insert the variable Smoke Cigarettes into the Rows box on the right.
Insert the variable Biological Sex into the Columns box on the right.
Choose OK.

This should result in the two-way table below:

Rows: Smokes Cigaretes | Columns: Biological Sex

	Female	Male	All
No	120	89	209
Yes	7	10	17
All	127	99	226
Cell Contents: Count

11.1.1 Probability Distribution Plots

In previous lessons you have constructed probabilities distribution plots for normal distributions, binomial distributions, and $t$ distributions. This week you will use the same procedure to construct a probability distribution plot for the chi-square distribution.

Minitab: Constructing a Chi-Square Probability Distribution Plot

Find a p-value from a chi-square test statistic in Minitab:

Select Graph > Probability Distribution Plot.
Select One Curve, then select View Probability.
Select OK.
Change the Distribution to Chi-Square.
Enter the Degrees of freedom.
Select Options.
Select A specified x value.
Select Right tail.
Under X value, enter the chi-square test statistic.
Select OK, then select OK again.

Example 11.1 (Minitab: Area to the Right of $\chi^2=1.75$) Chi-square tests of independence are always right-tailed tests. Let’s find the area of a chi-square distribution with 1 degree of freedom to the right of $\chi^2 = 1.75$. In other words, we’re looking up the $p$ value associated with a chi-square test statistic of 1.75.

In Minitab, select Graph > Probability Distribution Plot > View Probability.
For Distribution, select Chi-Square.
For Degrees of freedom, enter 1.
Select A specified X value.
Select Right tail.
For X value, enter 1.75.
Choose OK and OK.

This should result in the following output:

Chi-square distribution plot made using Minitab Express; degrees of freedom equal 1; the area to the right of a chi-square value of 1.75 is 0.185877. — Fig 11.1

Example 11.2 (Area to the Right of Chi-Sq = 6.25, df=3) Construct a chi-square distribution with 3 degrees of freedom to find the area to the right of a chi-square value of 6.25.

In Minitab, select Graph > Probability Distribution Plot > View Probability.
For Distribution, select Chi-Square.
For Degrees of freedom, enter 3.
Select A specified X value.
Select Right tail.
For X value, enter 6.25.
Choose OK and OK.

Chi-square distribution with the area shaded to the right of 6.25 — Fig 11.2

The area to the right of 6.25 in the chi-square distribution with 3 degrees of freedom is 0.1001.

11.1.2 Conditional Probabilities and Independence

In Lesson 2 you were introduced to conditional probabilities in conditional probabilities and independent events in independent events. These definitions are reviewed below, along with some examples.

Recall that if events A and B are independent then $P(A) = P(A \mid B)$. In other words, whether or not event B occurs does not change the probability of event A occurring.

Definition 11.3 (Conditional Probability) The probability of one event occurring given that it is known that a second event has occurred. This is communicated using the symbol $\mid$ which is read as “given.”

For example, $P(A\mid B)$ is read as “Probability of A given B.”

Definition 11.4 (Independent Events) Unrelated events. The outcome of one event does not impact the outcome of the other event.

Example 11.3 (Queens & Hearts) If a card is randomly drawn from a standard 52-card deck, the probability of the card being a queen is independent from the probability of the card being a heart. If I tell you that a randomly selected card is a queen, that does not change the likelihood of it being a heart, diamond, club, or spade.

Using a conditional probability to prove this:

$P(\text{Queen}) = \dfrac{4}{52}=0.077$

$P(\text{Queen} \mid \text{Heart}) = \dfrac {1}{13} = 0.077$

Example 11.4 (Gender and Pass Rate) Data concerning two categorical variables can be displayed in a contingency table.

	Pass	Did Not Pass	Total
Men	6	9	15
Women	10	15	25
Total	16	24	40

If gender and passing are independent, then the probability of passing will not change if a case’s gender is known. This could be written as $P(Pass) = P(Pass \mid Man)$.

$P(\text{Pass}) = \dfrac{16}{40} = 0.4$

$P(\text{Pass} \mid \text{Man}) = \dfrac{6}{15}=0.4$

In this sample, gender and passing are independent.

11.2 Goodness of Fit Test

A chi-square goodness-of-fit test can be conducted when there is one categorical variable with more than two levels. If there are exactly two categories, then a one proportion z test may be conducted. The levels of that categorical variable must be mutually exclusive. In other words, each case must fit into one and only one category.

We can test that the proportions are all equal to one another, or we can test any specific set of proportions.

If the expected counts, which we’ll learn how to compute shortly, are all at least five, then the chi-square distribution may be used to approximate the sampling distribution. If any expected count is less than five, then a randomization test should be conducted.

Possible Research Questions

According to one research study, about 90% of American adults are right-handed, 9% are left-handed, and 1% are ambidextrous. Are the proportions of Penn State students who are right-handed, left-handed, and ambidextrous different from these national values?
A concessions stand sells blue, red, purple, and green freezer pops. They survey a sample of children and ask which of the four colors is their favorite. They want to know if the colors differ in popularity.

Test Statistic

In conducting a goodness-of-fit test, we compare observed counts to expected counts. Observed counts are the number of cases in the sample in each group. Expected counts are computed given that the null hypothesis is true; this is the number of cases we would expect to see in each cell if the null hypothesis were true.

Expected Cell Value

$E=n (p_i)$

$n$ is the total sample size
$p_i$ is the hypothesized proportion of the “i^{th}” group

The observed and expected values are then used to compute the chi-square ($\chi^2$) test statistic.

Chi-Square ($\chi^2$) Test Statistic

\[ \chi^2=\sum \dfrac{(\text{Observed}-\text{Expected})^2}{\text{Expected}} \]

Approximating the Sampling Distribution

StatKey has the ability to conduct a randomization test for a goodness-of-fit test. There is an example of this in Section 7.1 of the Lock⁵ textbook. If all expected values are at least five, then the sampling distribution can be approximated using a chi-square distribution.

Like the t-distribution, the chi-square distribution varies depending on the degrees of freedom. Degrees of freedom for a chi-square goodness-of-fit test are equal to the number of groups minus 1. The distribution plot below compares the chi-square distributions with 2, 4, and 6 degrees of freedom.

Probability distribution plot made using Minitab Express; 3 chi-square distributions are overlaid with degrees of freedom of 2, 4, and 6. — Fig 11.3

To find the p-value we find the area under the chi-square distribution to the right of our test statistic. A chi-square test is always right-tailed.

11.2.1 Five Step Hypothesis Testing Procedure

The examples in the following section use the five step hypothesis testing procedure outlined below. This is the same procedure that we used to conduct a hypothesis test for a single mean, single proportion, difference in two means, and difference in two proportions.

Check assumptions and write hypotheses.

When conducting a chi-square goodness-of-fit test, it makes the most sense to write the hypotheses first. The hypotheses will depend on the research question. The null hypothesis will always contain the equalities and the alternative hypothesis will be that at least one population proportion is not as specified in the null.

In order to use the chi-square distribution to approximate the sampling distribution, all expected counts must be at least five.

Expected Count

$E=np_i$

Where $n$ is the total sample size and $p_i$ is the hypothesized population proportion in the “ith” group.

To check this assumption, compute all expected counts and confirm that each is at least five.
Compute the test statistic.

In Step 1 you already computed the expected counts. Use this formula to compute the chi-square test statistic:

Chi-Square Test Statistic

$\chi^2=\sum \dfrac{(O-E)^2}{E}$

Where $O$ is the observed count for each cell and $E$ is the expected count for each cell.
Determine the p-value.

Construct a chi-square distribution with degrees of freedom equal to the number of groups minus one. The p-value is the area under that distribution to the right of the test statistic that was computed in Step 2. You can find this area by constructing a probability distribution plot in Minitab.
Make a decision.

Unless otherwise stated, use the standard 0.05 alpha level.

$p \leq \alpha$ reject the null hypothesis.

$p > \alpha$ fail to reject the null hypothesis.
State a real-world conclusion.

Go back to the original research question and address it directly. If you rejected the null hypothesis, then there is convincing evidence that at least one of the population proportions is not as stated in the null hypothesis. If you failed to reject the null hypothesis, then there is not enough evidence that any of the population proportions are different from what is stated in the null hypothesis.

Example 11.5 (Cupcakes (Equal Proportions))

Video 11.1: STAT 200 ChiSquare Goodness of Fit Test: Cupcake Flavor Example

Example 11.6 (Cards (Equal Proportions)) Research question: When randomly selecting a card from a deck with replacement, are we equally likely to select a heart, diamond, spade, and club?

I randomly selected a card from a standard deck 40 times with replacement. I pulled 13 hearts, 8 diamonds, 8 spades, and 11 clubs.

Let’s use the five-step hypothesis testing procedure:

Check assumptions and write hypotheses.

$H_0: p_h=p_d=p_s=p_c=0.25$
$H_a:$ at least one $p_i$ is not as specified in the null

We can use the null hypothesis to check the assumption that all expected counts are at least 5.

Expected count $=n (p_i)$

All $p_i$ are 0.25. $40(0.25)=10$, thus this assumption is met and we can approximate the sampling distribution using the chi-square distribution.
Compute the test statistic.

$\chi^2=\sum \dfrac{(\text{Observed}-\text{Expected})^2}{\text{Expected}}$

All expected values are 10. Our observed values were 13, 8, 8, and 11.

$\chi^2=\dfrac{(13-10)^2}{10}+\dfrac{(8-10)^2}{10}+\dfrac{(8-10)^2}{10}+\dfrac{(11-10)^2}{10}$
$\chi^2=\dfrac{9}{10}+\dfrac{4}{10}+\dfrac{4}{10}+\dfrac{1}{10}$
$\chi^2=1.8$
Determine the p-value.

Our sampling distribution will be a chi-square distribution.

\[ df=k-1=4-1=3 \]

We can find the p-value by constructing a chi-square distribution with 3 degrees of freedom to find the area to the right of $\chi^2=1.8$

Fig 11.4

The p-value is 0.614935
Make a decision.

$p>0.05$ therefore we fail to reject the null hypothesis.
State a “real world” conclusion.

There is not enough evidence to state that the proportion of hearts, diamonds, spades, and clubs that are randomly drawn from this deck are different.

Example 11.7 (Roulette Wheel (Different Proportions)) Research Question: An American roulette wheel contains 38 slots: 18 red, 18 black, and 2 green. A casino has purchased a new wheel and they want to know if there is convincing evidence that the wheel is unfair. They spin the wheel 100 times and it lands on red 44 times, black 49 times, and green 7 times.

Check assumptions and write hypotheses.

If the wheel is fair then $p_{\text{red}}=\dfrac{18}{38}$, $p_{\text{black}}=\dfrac{18}{38}$, and $p_{\text{green}}=\dfrac{2}{38}$.

All of these proportions combined equal 1.

$H_0: p_{\text{red}}=\dfrac{18}{38}$, $p_{\text{black}}=\dfrac{18}{38}$, and $p_{\text{green}}=\dfrac{2}{38}$.

$H_a$: At least one $p_i$ is not as specified in the null.

In order to conduct a chi-square goodness of fit test all expected values must be at least 5.

For both red and black: Expected count $=100(\dfrac{18}{38})=47.368$.

For green: Expected count $=100(\dfrac{2}{38})=5.263$.

All expected counts are at least 5 so we can conduct a chi-square goodness of fit test.
Compute the test statistic.

$\chi^2=\sum \dfrac{(Observed-Expected)^2}{Expected}$

In the first step we computed the expected values for red and black to be 47.368 and for green to be 5.263.

$\chi^2= \dfrac{(44-47.368)^2}{47.368}+\dfrac{(49-47.368)^2}{47.368}+\dfrac{(7-5.263)^2}{5.263}$

$\chi^2=0.239+0.056+0.573=0.868$
Determine the p-value.

Our sampling distribution will be a chi-square distribution.

$df=k-1=3-1=2$

We can find the p-value by constructing a chi-square distribution with 2 degrees of freedom to find the area to the right of $\chi^2=0.868$

Fig 11.5

The p-value is 0.647912.
Make a decision.

$p>0.05$ therefore we should fail to reject the null hypothesis.
State a “real world” conclusion.

There is not enough evidence that this roulette wheel is unfair.

11.2.2 Minitab: Goodness-of-Fit Test

Conduct a chi-square goodness-of-fit test with summarized data in Minitab:

Enter the category names in one worksheet column and the observed counts in a second column.
Select Stat > Tables > Chi-Square Goodness of Fit Test (One Variable).
Under Observed counts, enter the column containing the observed counts.
Under Category names (optional), enter the column containing the category names.
To test whether all population proportions are equal, use the default option for equal proportions.
Select OK.

Example 11.8 (Minitab: Cards (Equal Proportions)) Research Question: When randomly selecting a card from a deck with replacement, are we equally likely to select a heart, diamond, spade, and club?

I randomly selected a card from a standard deck 40 times with replacement. I pulled 13 Hearts (♥), 8 Diamonds (♦), 8 Spades (♠), and 11 Clubs (♣).

To perform a chi-square goodness-of-fit test in Minitab using summarized data we first need to enter the data into the worksheet. Below you can see that we have one column with the names of each group and one column with the observed counts for each group.

	C1	C2
	Suit	Count
1	Hearts	13
2	Diamonds	8
3	Spades	8
4	Clubs	11

After entering the data, select Stat > Tables > Chi-Square Goodness of Fit Test (One Variable).
Insert Count into the Observed Counts box.
Insert Suit into the Category names (optional) box.
Choose OK.

This should result in the following output:

Chi-Square Goodness-of-Fit Test: Count

Observed and Expected Counts

Category	Observed	TestProportion	Expected	Contributionto Chi-Sq
Hearts	13	0.25	10	0.9
Diamonds	8	0.25	10	0.4
Spades	8	0.25	10	0.4
Clubs	11	0.25	10	0.1

Chi-Square Test

N	DF	Chi-Sq	P-Value
40	3	1.8	0.615

All expected values are at least 5 so we can use the chi-square distribution to approximate the sampling distribution. Our results are $\chi^2 (3) = 1.8$. $p = 0.615$. Because our p-value is greater than the standard alpha level of 0.05, we fail to reject the null hypothesis. There is not enough evidence to conclude that the proportions are different in the population.

Note!

The example above tested equal population proportions. Minitab also has the ability to conduct a chi-square goodness-of-fit test when the hypothesized population proportions are not all equal. To do this, you can choose to test specified proportions or to use proportions based on historical counts.

Example 11.9 (Tulips (Summarized Data, Equal Proportions)) A company selling tulip bulbs claims they have equal proportions of white, pink, and purple bulbs and that they fill customer orders by randomly selecting bulbs from the population of all of their bulbs.

You ordered 30 bulbs and received 16 white, 8 pink, and 6 purple.

Is there convincing evidence the bulbs you received were not randomly selected from a population with an equal proportion of each color?

Use Minitab to conduct a hypothesis test to address this research question.

We’ll go through each of the steps in the hypotheses test:

Check assumptions and write hypotheses.

$H_0\colon p_{white}=p_{pink}=p_{purple}=\dfrac{1}{3}$
$H_a\colon$ at least one $p_i$ is not $\dfrac{1}{3}$

We can use the null hypothesis to check the assumption that all expected counts are at least 5.

$Expected\;count=n (p_i)$

All $p_i$ are $\frac{1}{3}$. $30(\frac{1}{3})=10$, thus this assumption is met and we can approximate the sampling distribution using the chi-square distribution.
Compute the test statistic.

Let’s use Minitab to calculate this.

First, enter the summarized data into a Minitab Worksheet.

C1 C2

Color Count

1 White 16

2 Pink 8

3 Purple 6
1. After entering the data, select Stat > Tables > Chi-Square Goodness of Fit Test (One Variable).
2. Insert Count into the Observed Counts box.
3. Insert Color into the Category names (optional) box.
4. Choose OK.
This should result in the following output:

Chi-Square Goodness-of-Fit Test: Count

Observed and Expected Counts

Category Observed Test Proportion Expected Contribution to Chi-Sq

White 16 0.333333 10 3.6

Pink 8 0.333333 10 0.4

Purple 6 0.333333 10 1.6

Chi-Square Test

N DF Chi-Sq P-Value

30 2 5.6 0.061

The test statistic is a Chi-Square of 5.6.
Determine the p-value.

The p-value from the output is 0.061.
Make a decision.

$p>0.05$ therefore we fail to reject the null hypothesis.
State a “real world” conclusion.

There is not enough evidence that the tulip bulbs were not randomly selected from a population with equal proportions of white, pink and purple.

	C1	C2
	Color	Count
1	White	16
2	Pink	8
3	Purple	6

Category	Observed	Test Proportion	Expected	Contribution to Chi-Sq
White	16	0.333333	10	3.6
Pink	8	0.333333	10	0.4
Purple	6	0.333333	10	1.6

N	DF	Chi-Sq	P-Value
30	2	5.6	0.061

Example 11.10 (Roulette (Summarized Data, Different Proportions)) An American roulette wheel contains 38 slots: 18 red, 18 black, and 2 green. A casino has purchased a new wheel and they want to know if there is convincing evidence that the wheel is unfair. They spin the wheel 100 times and it lands on red 44 times, black 49 times, and green 7 times.

Use Minitab to conduct a hypothesis test to address this question.

We’ll go through each of the steps in the hypotheses test:

Check assumptions and write hypotheses.

If the wheel is ‘fair’ then the probability of red and black are both 18/38 and the probability of green is 2/38.

$H_0\colon p_{red}=\dfrac{18}{38}, p_{black}=\dfrac{18}{38}, p_{green}=\dfrac{2}{38}$
$H_a\colon$ at least one $p_i$ is not as specified in the null

We can use the null hypothesis to check the assumption that all expected counts are at least 5.

$\text{Expected count}=n (p_i)$

With n = 100 we meet the assumptions needed to use Chi-square.

Compute the test statistic.

Let’s use Minitab to calculate this.

First, enter the summarized data into a Minitab Worksheet.

	C1	C2	C3
	Like Tea	Like Coffee—Yes	Like Coffee—No
1	Yes	30	25
2	No	10	35

After entering the data, select Stat > Tables > Chi-Square Goodness of Fit Test (One Variable).
Insert Count into the Observed Counts box.
Insert Color into the Category names (optional) box.
For Test, select Input constants.
Select Proportions specified by historical counts (this is what we would expect if the null was true).
Enter 18/38 for Black, 2/38 for Green, and 18/38 for Red.
Choose OK.

This should result in the following output:

Chi-Square Goodness-of-Fit Test: Count

Observed and Expected Counts

Category	Observed	Historical Counts	Test Proportion	Expected	Contribution to Chi-Sq
Red	44	18	0.473684	47.3684	0.239532
Black	49	18	0.473684	47.3684	0.056199
Green	7	2	0.052632	5.2632	0.573158

Chi-Square Test

N	DF	Chi-Sq	P-Value
100	2	0.868889	0.648

The test statistic is a Chi-Square of 0.87.

Determine the p-value.

The p-value from the output is 0.648.
Make a decision.

$p>0.05$ therefore we fail to reject the null hypothesis.
State a “real world” conclusion.

There is not enough evidence to state that this roulette wheel is unfair.

11.3 Chi-Square Test of Independence

The chi-square ($\chi^2$) test of independence is used to test for a relationship between two categorical variables. Recall that if two categorical variables are independent, then $P(A) = P(A \mid B)$. The chi-square test of independence uses this fact to compute expected values for the cells in a two-way contingency table under the assumption that the two variables are independent (i.e., the null hypothesis is true).

Even if two variables are independent in the population, samples will vary due to random sampling variation. The chi-square test is used to determine if there is convincing evidence that the two variables are not independent in the population using the same hypothesis testing logic that we used with one mean, one proportion, etc.

Again, we will be using the five-step hypothesis testing procedure:

Check assumptions and write hypotheses.

The assumptions are that the sample is randomly drawn from the population and that all expected values are at least 5 (we will see what expected values are later).

Our hypotheses are:

$H_0:$ There is not a relationship between the two variables in the population (they are independent)

$H_a:$ There is a relationship between the two variables in the population (they are dependent)

Note: When you’re writing the hypotheses for a given scenario, use the names of the variables, not the generic “two variables.”
Calculate an appropriate test statistic.

Chi-Square Test Statistic

\[ \chi^2=\sum \dfrac{(\text{Observed}-\text{Expected})^2}{\text{Expected}} \]

Expected Cell Value

\[ E=\dfrac{\text{row total} \times \text{column total}}{n} \]
Determine the p-value.

The p-value can be found using Minitab. Look up the area to the right of your chi-square test statistic on a chi-square distribution with the correct degrees of freedom. Chi-square tests are always right-tailed tests.

Degrees of Freedom: Chi-Square Test of Independence

$df=(\text{number of rows}-1)(\text{number of columns}-1)$
Make a decision.

If $p \leq \alpha$ reject the null hypothesis.

If $p>\alpha$ fail to reject the null hypothesis.
State a “real world” conclusion.

Write a conclusion in terms of the original research question.

Example 11.11 (Gender and Online Learning) A sample of 314 Penn State students was asked if they have ever taken an online course. Their genders were also recorded. The contingency table below was constructed. Use a chi-square test of independence to determine if there is a relationship between gender and whether or not someone has taken an online course.

Have you taken an online course?
	Yes	No
Men	43	63
Women	95	113

Solution

Check assumptions and write hypotheses.

$H_0:$ There is not a relationship between gender and whether or not someone has taken an online course (they are independent)

$H_a:$ There is a relationship between gender and whether or not someone has taken an online course (they are dependent)

Looking ahead to our calculations of the expected values, we can see that all expected values are at least 5. This means that the sampling distribution can be approximated using the $\chi^2$ distribution.
Calculate the test statistic.

In order to compute the chi-square test statistic we must know the observed and expected values for each cell. We are given the observed values in the table above. We must compute the expected values. The table below includes the row and column totals.

Have you taken an online course?

Yes No Total

Men 43 63 106

Women 95 113 208

Total 138 176 314

$E=\dfrac{row\;total \times column\;total}{n}$
$E_{\text{Men, Yes}}=\dfrac{106\times138}{314}=46.586$
$E_{\text{Men, No}}=\dfrac{106\times176}{314}=59.414$
$E_{\text{Women, Yes}}=\dfrac{208\times138}{314}=91.414$
$E_{\text{Women, No}}=\dfrac{208 \times 176}{314}=116.586$

Note that all expected values are at least 5, thus this assumption of the $\chi^2$ test of independence has been met.

Observed and expected counts are often presented together in a contingency table. In the table below, expected values are presented in parentheses.

Have you taken an online course?

Yes No Total

Men 43 (46.586) 63 (59.414) 106

Women 95 (91.414) 113 (116.586) 208

Total 138 176 314

\[ \chi^2=\sum \dfrac{(O-E)^2}{E} \]

\[ \begin{align*} \chi^2&=\dfrac{(43-46.586)^2}{46.586}+\dfrac{(63-59.414)^2}{59.414}+\dfrac{(95-91.414)^2}{91.414}+\dfrac{(113-116.586)^2}{116.586} \\ &=0.276+0.216+0.141+0.110 \\ &=0.743 \end{align*} \]

The chi-square test statistic is 0.743.

\[ df=(\text{number of rows}-1)(\text{number of columns}-1)=(2-1)(2-1)=1 \]
Determine the p-value.

We can determine the p-value by constructing a chi-square distribution plot with 1 degree of freedom and finding the area to the right of 0.743.

Fig 11.6

$p = 0.388702$
Make a decision.

$p>\alpha$, therefore we fail to reject the null hypothesis.
State a “real world” conclusion.

There is not enough evidence to conclude that gender and whether or not an individual has completed an online course are related.

Note that we cannot say for sure that these two categorical variables are independent, we can only say that we do not have enough evidence that they are dependent.

Have you taken an online course?
	Yes	No	Total
Men	43	63	106
Women	95	113	208
Total	138	176	314

Have you taken an online course?
	Yes	No	Total
Men	43 (46.586)	63 (59.414)	106
Women	95 (91.414)	113 (116.586)	208
Total	138	176	314

11.3.1 Minitab: Test of Independence

Raw vs Summarized Data

If you have a data file with the responses for individual cases then you have “raw data” and can follow the directions below. If you have a table filled with data, then you have “summarized data.” There is an example of conducting a chi-square test of independence using summarized data in a later section. After data entry the procedure is the same for both data entry methods.

Minitab: Chi-square Test Using Raw Data

Conduct a chi-square test of association in Minitab:

Select Stat > Tables > Chi-Square Test for Association.
Select the appropriate data option for raw or summarized data.
Select the categorical variables.
Select the Statistics tab.
Select the statistics you want Minitab to display.
Select OK.

Example 11.12 (Minitab: Seat Location and Cheating) Research question: Is there a relationship between where a student sits in class and whether they have ever cheated?

Null hypothesis: Seat location and cheating are not related in the population.
Alternative hypothesis: Seat location and cheating are related in the population.

To perform a chi-square test of independence in Minitab using raw data:

Open Minitab file: (class_survey.mpx)
Select Stat > Tables > Chi-Square Test for Association.
Select Raw data (categorical variables) from the dropdown menu.
Insert the variable Seating into the Rows box.
Insert the variable Ever_Cheat into the Columns box.
Select Statistics and check the boxes Chi-square test for association and Expected cell counts.
Choose OK and OK.

This should result in the following output:

Rows: Seating Columns: Ever_Cheat

	No	Yes	All
Back	24 24.21	8 7.79	32
Front	38 34.81	8 11.19	46
Middle	109 111.98	39 36.02	148
All	1714	55	226

Cell contents: Count
Expected count

Chi-Square Test

	Chi-Square	DF	P-Value
Pearson	1.539	2	0.463
Likelihood Ratio	1.626	2	0.443

Interpret

All expected values are at least 5 so we can use the Pearson chi-square test statistic. Our results are $\chi^2 (2) = 1.539$. $p = 0.463$. Because our $p$ value is greater than the standard alpha level of 0.05, we fail to reject the null hypothesis. There is not enough evidence of a relationship in the population between seat location and whether a student has cheated.

Example 11.13 (Dog & Cat Ownership (Raw Data)) Is there a relationship between dog and cat ownership in the population of all World Campus STAT 200 students? Let’s conduct a hypothesis test using the dataset: (fall2016stdata.mpx)

Check any necessary assumptions and write null and alternative hypotheses.

$H_0:$ There is not a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students
$H_a:$ There is a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students

Assumption: All expected counts are at least 5. The expected counts here are 176.02, 75.98, 189.98, and 82.02, so this assumption has been met.
Calculate an appropriate test statistic.

Let’s use Minitab to calculate the test statistic and p-value.
1. After entering the data, select Stat > Tables > Cross Tabulation and Chi-Square.
2. Enter Dog into the Rows box.
3. Enter Cat in the Columns box.
4. Select the Chi-Square button, and in the new window, check the box for the Chi-square test and Expected cell counts.
5. Choose OK and OK.
Rows: Dog Columns: Cat

No Yes All

No 183
176.02 69
75.98 252

Yes 183
189.98 89
82.02 272

Missing 1 0

All 366 158 524

Cell contents:

Count
Expected count

Chi-Square Test

Chi-Square DF P-Value

Pearson 1.771 1 0.183

Likelihood Ratio 1.775 1 0.183

Since the assumption was met in step 1, we can use the Pearson chi-square test statistic.

$\text{Pearson's}\chi^2 = 1.771$
Determine a p-value associated with the test statistic.

$p = 0.183$
Decide between the null and alternative hypotheses.

Our p-value is greater than the standard 0.05 alpha level, so we fail to reject the null hypothesis.
State a “real world” conclusion.

There is not enough evidence of a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students.

	No	Yes	All
No	183 176.02	69 75.98	252
Yes	183 189.98	89 82.02	272
Missing	1	0
All	366	158	524

	Chi-Square	DF	P-Value
Pearson	1.771	1	0.183
Likelihood Ratio	1.775	1	0.183

Example 11.14 (Coffee and Tea Preference (Summarized Data)) Is there a relationship between liking tea and liking coffee?

The following table shows data collected from a random sample of 100 adults. Each were asked if they liked coffee (yes or no) and if they liked tea (yes or no).

Coffee and tea preferences
Tea preference	Likes coffee	Does not like coffee
Likes tea	30	25
Does not like tea	10	35

Let’s use the five-step hypothesis testing procedure to address this research question.

Check any necessary assumptions and write null and alternative hypotheses.

$H_0:$ Liking coffee and liking tea are not related (i.e., independent) in the population
$H_a:$ Liking coffee and liking tea are related (i.e., dependent) in the population

Assumption: All expected counts are at least 5.
Calculate an appropriate test statistic.

Let’s use Minitab to calculate the test statistic and p-value.
1. Enter the table into a Minitab worksheet as shown below:
C1 C2 C3

Likes Tea Likes Coffee-Yes Likes Coffee-No

1 Yes 30 25

2 No 10 35
1. Select Stat > Tables > Cross Tabulation and Chi-Square.
2. Select Summarized data in a two-way table from the dropdown.
3. Enter the columns Likes Coffee-Yes and Likes Coffee-No in the Columns containing the table box.
4. For the row labels, enter Likes Tea (leave the column labels blank).
5. Select the Chi-Square button and check the boxes for Chi-square test and Expected cell counts.
6. Choose OK and OK.
Output

Rows: Likes Tea Columns: Worksheet columns

No Yes All

Yes 30
22 25
33 55

No 10
18 35
27 45

All 40 60 100

Cell contents:

Count
Expected count

Chi-Square Test

Chi-Square DF P-Value

Pearson 10.774 1 0.001

Likelihood Ratio 11.138 1 0.001

Since the assumption was met in step 1, we can use the Pearson chi-square test statistic.

$\text{Pearson's }\chi^2 = 10.774$
Determine a p-value associated with the test statistic.

$p = 0.001$
Decide between the null and alternative hypotheses.

Our p-value is less than the standard 0.05 alpha level, so we reject the null hypothesis.
State a “real world” conclusion.

There is convincing evidence of a relationship between liking coffee and liking tea in the population.

	C1	C2	C3
	Likes Tea	Likes Coffee-Yes	Likes Coffee-No
1	Yes	30	25
2	No	10	35

	No	Yes	All
Yes	30 22	25 33	55
No	10 18	35 27	45
All	40	60	100

	Chi-Square	DF	P-Value
Pearson	10.774	1	0.001
Likelihood Ratio	11.138	1	0.001

11.3.2 Relative Risk

A chi-square test of independence will give you information concerning whether or not a relationship between two categorical variables in the population is likely. As was the case with the single sample and two sample hypothesis tests that you learned earlier this semester, with a large sample size statistical power is high and the probability of rejecting the null hypothesis is high, even if the relationship is relatively weak. In addition to examining statistical significance by looking at the p-value, we can also examine practical significance by computing the relative risk.

In Lesson 2 you learned that risk is often used to describe the probability of an event occurring. Risk can also be used to compare the probabilities in two different groups. First, we’ll review risk, then you’ll be introduced to the concept of relative risk.

Definition 11.5 (Risk) The risk of an outcome can be expressed as a fraction or as the percent of a group that experiences the outcome. \[ \text{Risk}=\dfrac{\text{n with outcome}}{\text{total n}} \]

Example 11.15 (Examples of Risk) Asthma:
60 out of 1000 teens have asthma. The risk is $\frac{60}{1000}=.06$. This means that 6% of all teens experience asthma.

Flu:
45 out of 100 children get the flu each year. The risk is $\frac{45}{100}=.45$ or 45%

Definition 11.6 (Relative Risk) Relative risk compares the risk of a particular outcome in two different groups.

\[ \text{Relative Risk}=\dfrac{\text{Risk in Group 1}}{\text{Risk in Group 2}} \]

Thus, relative risk gives the risk for group 1 as a multiple of the risk for group 2.

Example 11.16 (Relative Risk for Flu in Children and Adults) Suppose that the risk of a child getting the flu this year is .45 and the risk of an adult getting the flu this year is .10. What is the relative risk of children compared to adults?

\[ \text{Relative risk}=\dfrac{.45}{.10}=4.5 \]

Children are 4.5 times more likely than adults to get the flu this year.

Cautions

Watch out for relative risk statistics where no baseline information is given about the actual risk. For instance, it doesn’t mean much to say that beer drinkers have twice the risk of stomach cancer as non-drinkers unless we know the actual risks. The risk of stomach cancer might actually be very low, even for beer drinkers. For example, 2 in a million is twice the size of 1 in a million but is would still be a very low risk. This is known as the baseline with which other risks are compared.

11.4 Summary

In this lesson you added two new statistical procedures to your repertoire: the chi-square goodness-of-fit test for one categorical variable and the chi-square test of independence for two variables. In the remaining weeks we will continue to learn more procedures.

--- categories: [Categorical Data, Chi-Square Tests, Goodness-of-Fit Tests, Tests of Independence, Relative Risk] image: /assets/L11card.png # change the image path --- # Chi-Square Tests ::: objectiveblock ## <i class="bi bi-check2-circle"></i>Objectives {.unnumbered .unlisted} Upon successful completion of this lesson, you should be able to: 1. Construct a chi-square probability distribution plot in StatKey or Minitab. 2. Determine when a chi-square goodness-of-fit test and chi-square test of independence should be conducted. 3. Compute and interpret expected counts. 4. Conduct chi-square tests by hand and using Minitab. 5. Calculate and interpret relative risk. ::: []{.bi .bi-book .fs-4} This lesson corresponds to Chapter 7 in the Lock<sup>5</sup> textbook. In this lesson we will learn how to compare the proportions of more than two independent groups and how to test for a relationship between two categorical variables. In this lesson you will also review risk and odds and learn about relative risk. Before we begin, let's review the various classifications of variables from [Lesson 1](Lesson01.qmd). ::: {#def-categorical} ### Categorical Names or labels (i.e., categories) with no logical order or with a logical order but inconsistent differences between groups. Also known as qualitative. ::: ::: {#def-quantitative} ### Quantitative {.unnumbered .unlisted} Numerical values with magnitudes that can be placed in a meaningful order with consistent intervals. Also known as numerical. ::: In this lesson, we will be examining methods for analyzing **categorical** variables. ## Review In this lesson you may need to use Minitab to construct frequency tables or two-way contingency tables. We'll start by reviewing these procedures. You will need to construct probability distribution plots for chi-square distributions. In earlier lessons we constructed probability distribution plots for z-, t-, and F-distributions; the procedure is similar for a chi-square distribution. We will also review conditional probabilities and the term *independence* in this section. ### Frequency Tables { .unnumbered} The following example was first presented in [Lesson 2](Lesson02.qmd#sec-minitab-frequency-table). It uses following data set (from College Board): ([sat_data.mpx](data_files/sat_data.mpx){download="" target="_blank"}) #### ![ ](assets/minitab.png){.icon-image} Minitab: Frequency Tables ::: {.ms-3 .border-start .border-secondary-subtle .ps-3} To create a frequency table in Minitab: 1. Open the Minitab file: ([sat_data.mpx](data_files/sat_data.mpx){download="" target="_blank"}) 2. Select **Stat > Tables > Tally Individual Variables**. 3. Insert the variable *Region* into the **Variable** box on the right. 4. Under **Statistics**, check **Counts** and **Percents**. 5. Choose **OK**. This should result in the following frequency table: #### Tally {.unnumbered .unlisted} ::: {.minitab_output} | Region | Count | Percent | | --- |--- |--- | | ENC | 5 | 9.80 | | ESC | 4 | 7.84 | | MA | 3 | 5.88 | | MTN | 8 | 15.69 | | NE | 6 | 11.76 | | PAC | 5 | 9.80 | | SA | 9 | 17.65 | | WNC | 7 | 13.73 | | WSC | 4 | 7.84 | | N= | 51 | | : {.w-auto .table-sm .row-header} ::: ::: ### Two-Way Contingency Table {.unnumbered} Recall from @sec-two-cat-var that a **two-way contingency table** is a display of counts for two categorical variables in which the rows represented one variable and the columns represent a second variable. The starting point for analyzing the relationship between two categorical variables is to create a two-way contingency table. When one variable is obviously the explanatory variable, the convention is to use the explanatory variable to define the rows and the response variable to define the columns; this is not a hard and fast rule though. #### ![ ](assets/minitab.png){.icon-image} Minitab: Constructing a Two-Way Contingency Table ::: {.ms-3 .border-start .border-secondary-subtle .ps-3} 1. Open the data set: ([class_survey.mpx](data_files/class_survey.mpx){download="" target="_blank"}) 2. Select **Stat > Tables > Cross Tabulation and Chi-square**. 3. Select **Raw data (categorical variable)** from the dropdown menu. 4. Insert the variable *Smoke Cigarettes* into the **Rows** box on the right. 5. Insert the variable *Biological Sex* into the **Columns** box on the right. 6. Choose **OK**. This should result in the two-way table below: ::: {.minitab_output} #### Rows: Smokes Cigaretes | Columns: Biological Sex {.unnumbered .unlisted} | | Female | Male | All | | --- |--- |--- |--- | | No | 120 | 89 | 209 | | Yes | 7 | 10 | 17 | | All | 127 | 99 | 226 | | Cell Contents: Count | : {.w-auto .table-sm .row-header} ::: ::: ### Probability Distribution Plots In previous lessons you have constructed probabilities distribution plots for normal distributions, binomial distributions, and $t$ distributions. This week you will use the same procedure to construct a probability distribution plot for the chi-square distribution. #### ![ ](assets/minitab.png){.icon-image} Minitab: Constructing a Chi-Square Probability Distribution Plot {{< include _minitab/find-a-p-value-from-a-chi-square-test-statistic.qmd >}} ::: {#exm-area-to-the-right-of-chi-square-1-75-in-minitab .example-minitab} #### Minitab: Area to the Right of $\chi^2=1.75$ Chi-square tests of independence are always right-tailed tests. Let's find the area of a chi-square distribution with 1 degree of freedom to the right of $\chi^2 = 1.75$. In other words, we're looking up the $p$ value associated with a chi-square test statistic of 1.75. 1. In Minitab, select **Graph > Probability Distribution Plot > View Probability**. 2. For **Distribution**, select **Chi-Square**. 3. For **Degrees of freedom**, enter 1. 4. Select **A specified X value**. 5. Select **Right tail**. 6. For **X value**, enter 1.75. 7. Choose **OK** and **OK**. This should result in the following output: ![](assets/200chisquex.png){#fig-200chisquex fig-alt="Chi-square distribution plot made using Minitab Express; degrees of freedom equal 1; the area to the right of a chi-square value of 1.75 is 0.185877." .mx-auto .d-block width="60%" .lightbox} ::: ::: {#exm-area-to-the-right-of-chi} #### Area to the Right of Chi-Sq = 6.25, df=3 Construct a chi-square distribution with 3 degrees of freedom to find the area to the right of a chi-square value of 6.25. 1. In Minitab, select **Graph > Probability Distribution Plot > View Probability**. 2. For **Distribution**, select **Chi-Square**. 3. For **Degrees of freedom**, enter 3. 4. Select **A specified X value**. 5. Select **Right tail**. 6. For **X value**, enter 6.25. 7. Choose **OK** and **OK**. ![](assets/200chisquare2ndex.png){#fig-200chisquare2ndex fig-alt="Chi-square distribution with the area shaded to the right of 6.25" .mx-auto .d-block width="60%" .lightbox} The area to the right of 6.25 in the chi-square distribution with 3 degrees of freedom is 0.1001. ::: ### Conditional Probabilities and Independence In [Lesson 2](Lesson02.qmd) you were introduced to conditional probabilities in [conditional probabilities](Lesson02.qmd#sec-conditional-probability) and independent events in [independent events](Lesson02.qmd#sec-disjoint-and-independent-events). These definitions are reviewed below, along with some examples. Recall that if events A and B are independent then $P(A) = P(A \mid B)$. In other words, whether or not event B occurs does not change the probability of event A occurring. ::: {#def-conditionalprobability} #### Conditional Probability The probability of one event occurring given that it is known that a second event has occurred. This is communicated using the symbol $\mid$ which is read as "given." For example, $P(A\mid B)$ is read as "Probability of A given B." ::: ::: {#def-independentevents} #### Independent Events Unrelated events. The outcome of one event does not impact the outcome of the other event. ::: ::: {#exm-queens-hearts} #### Queens & Hearts If a card is randomly drawn from a standard 52-card deck, the probability of the card being a queen is independent from the probability of the card being a heart. If I tell you that a randomly selected card is a queen, that does not change the likelihood of it being a heart, diamond, club, or spade. Using a conditional probability to prove this: $P(\text{Queen}) = \dfrac{4}{52}=0.077$ $P(\text{Queen} \mid \text{Heart}) = \dfrac {1}{13} = 0.077$ ::: ::: {#exm-gender-and-pass-rate} #### Gender and Pass Rate Data concerning two categorical variables can be displayed in a contingency table. | | Pass | Did Not Pass | Total | | --- |--- |--- |--- | | Men | 6 | 9 | 15 | | Women | 10 | 15 | 25 | | Total | 16 | 24 | 40 | : {.w-auto .table-sm .mx-auto .row-header} If gender and passing are independent, then the probability of passing will not change if a case's gender is known. This could be written as $P(Pass) = P(Pass \mid Man)$. $P(\text{Pass}) = \dfrac{16}{40} = 0.4$ $P(\text{Pass} \mid \text{Man}) = \dfrac{6}{15}=0.4$ In this sample, gender and passing are independent. ::: ## Goodness of Fit Test A chi-square goodness-of-fit test can be conducted when there is one categorical variable with more than two levels. If there are exactly two categories, then a one proportion z test may be conducted. The levels of that categorical variable must be mutually exclusive. In other words, each case must fit into one and only one category. We can test that the proportions are all equal to one another, or we can test any specific set of proportions. If the expected counts, which we'll learn how to compute shortly, are all at least five, then the chi-square distribution may be used to approximate the sampling distribution. If any expected count is less than five, then a randomization test should be conducted. ### Possible Research Questions {.unnumbered .unlisted} - According to one research study, about 90% of American adults are right-handed, 9% are left-handed, and 1% are ambidextrous. Are the proportions of Penn State students who are right-handed, left-handed, and ambidextrous different from these national values? - A concessions stand sells blue, red, purple, and green freezer pops. They survey a sample of children and ask which of the four colors is their favorite. They want to know if the colors differ in popularity. ### Test Statistic {.unnumbered .unlisted} In conducting a goodness-of-fit test, we compare observed counts to expected counts. Observed counts are the number of cases in the sample in each group. Expected counts are computed given that the null hypothesis is true; this is the number of cases we would expect to see in each cell if the null hypothesis were true. ::: {.callout-note appearance="minimal"} [**Expected Cell Value**]{.fs-4} $E=n (p_i)$ $n$ is the total sample size $p_i$ is the hypothesized proportion of the "i^{th}" group ::: The observed and expected values are then used to compute the chi-square ($\chi^2$) test statistic. ::: {.callout-note appearance="minimal"} [**Chi-Square ($\chi^2$) Test Statistic**]{.fs-4} $$ \chi^2=\sum \dfrac{(\text{Observed}-\text{Expected})^2}{\text{Expected}} $$ ::: ### Approximating the Sampling Distribution {.unnumbered .unlisted} [StatKey](http://www.lock5stat.com/StatKey/) has the ability to conduct a randomization test for a goodness-of-fit test. There is an example of this in Section 7.1 of the Lock<sup>5</sup> textbook. If all expected values are at least five, then the sampling distribution can be approximated using a chi-square distribution. Like the t-distribution, the chi-square distribution varies depending on the degrees of freedom. Degrees of freedom for a chi-square goodness-of-fit test are equal to the number of groups minus 1. The distribution plot below compares the chi-square distributions with 2, 4, and 6 degrees of freedom. ![](assets/ChiSqDistributions.png){#fig-ChiSqDistributions fig-alt="Probability distribution plot made using Minitab Express; 3 chi-square distributions are overlaid with degrees of freedom of 2, 4, and 6." .mx-auto .d-block width="60%" .lightbox} To find the p-value we find the area under the chi-square distribution to the right of our test statistic. A chi-square test is always right-tailed. ### Five Step Hypothesis Testing Procedure The examples in the following section use the five step hypothesis testing procedure outlined below. This is the same procedure that we used to conduct a hypothesis test for a single mean, single proportion, difference in two means, and difference in two proportions. ::: {.steps} 1. **Check assumptions and write hypotheses.** When conducting a chi-square goodness-of-fit test, it makes the most sense to write the hypotheses first. The hypotheses will depend on the research question. The null hypothesis will always contain the equalities and the alternative hypothesis will be that at least one population proportion is not as specified in the null. In order to use the chi-square distribution to approximate the sampling distribution, all expected counts must be at least five. ::: {.callout-note appearance="minimal"} [**Expected Count**]{.fs-4} $E=np_i$ Where $n$ is the total sample size and $p_i$ is the hypothesized population proportion in the "ith" group. ::: To check this assumption, compute all expected counts and confirm that each is at least five. 2. **Compute the test statistic.** In Step 1 you already computed the expected counts. Use this formula to compute the chi-square test statistic: ::: {.callout-note appearance="minimal"} [**Chi-Square Test Statistic**]{.fs-4} $\chi^2=\sum \dfrac{(O-E)^2}{E}$ Where $O$ is the observed count for each cell and $E$ is the expected count for each cell. ::: 3. **Determine the p-value.** Construct a chi-square distribution with degrees of freedom equal to the number of groups minus one. The p-value is the area under that distribution to the right of the test statistic that was computed in Step 2. You can find this area by constructing a probability distribution plot in Minitab. 4. **Make a decision.** Unless otherwise stated, use the standard 0.05 alpha level. $p \leq \alpha$ reject the null hypothesis. $p > \alpha$ fail to reject the null hypothesis. 5. **State a real-world conclusion.** Go back to the original research question and address it directly. If you rejected the null hypothesis, then there is convincing evidence that at least one of the population proportions is not as stated in the null hypothesis. If you failed to reject the null hypothesis, then there is not enough evidence that any of the population proportions are different from what is stated in the null hypothesis. ::: ::: {#exm-cupcakes-equal-proportions} #### Cupcakes (Equal Proportions) ::: {.w-75 .mx-auto #vid-11-1} {{< video https://youtu.be/ZGYBZ0aUh7Q?si=tB0m-JvF2rftMR8Z >}} STAT 200 ChiSquare Goodness of Fit Test: Cupcake Flavor Example ::: ::: ::: {#exm-cards1} #### Cards (Equal Proportions) **Research question**: When randomly selecting a card from a deck with replacement, are we equally likely to select a heart, diamond, spade, and club? I randomly selected a card from a standard deck 40 times with replacement. I pulled 13 hearts, 8 diamonds, 8 spades, and 11 clubs. Let's use the five-step hypothesis testing procedure: ::: {.steps} 1. **Check assumptions and write hypotheses.** $H_0: p_h=p_d=p_s=p_c=0.25$ $H_a:$ at least one $p_i$ is not as specified in the null We can use the null hypothesis to check the assumption that all expected counts are at least 5. Expected count $=n (p_i)$ All $p_i$ are 0.25. $40(0.25)=10$, thus this assumption is met and we can approximate the sampling distribution using the chi-square distribution. 2. **Compute the test statistic.** $\chi^2=\sum \dfrac{(\text{Observed}-\text{Expected})^2}{\text{Expected}}$ All expected values are 10. Our observed values were 13, 8, 8, and 11. $\chi^2=\dfrac{(13-10)^2}{10}+\dfrac{(8-10)^2}{10}+\dfrac{(8-10)^2}{10}+\dfrac{(11-10)^2}{10}$ $\chi^2=\dfrac{9}{10}+\dfrac{4}{10}+\dfrac{4}{10}+\dfrac{1}{10}$ $\chi^2=1.8$ 3. **Determine the p-value.** Our sampling distribution will be a chi-square distribution. $$ df=k-1=4-1=3 $$ We can find the p-value by constructing a chi-square distribution with 3 degrees of freedom to find the area to the right of $\chi^2=1.8$ ![](assets/chisqcardspvalue.png){#fig-chisqcardspvalue fig-alt="Chi-squared distribution plot made using Minitab Express; degrees of freedom equal 3; area to the right of chi-squared value of 1.8 is 0.614935" .mx-auto .d-block width="60%" .lightbox} The p-value is 0.614935 4. **Make a decision.** $p>0.05$ therefore we fail to reject the null hypothesis. 5. **State a "real world" conclusion.** There is not enough evidence to state that the proportion of hearts, diamonds, spades, and clubs that are randomly drawn from this deck are different. ::: ::: ::: {#exm-roulette-wheel} #### Roulette Wheel (Different Proportions) **Research Question**: An American roulette wheel contains 38 slots: 18 red, 18 black, and 2 green. A casino has purchased a new wheel and they want to know if there is convincing evidence that the wheel is unfair. They spin the wheel 100 times and it lands on red 44 times, black 49 times, and green 7 times. ::: {.steps} 1. **Check assumptions and write hypotheses.** If the wheel is fair then $p_{\text{red}}=\dfrac{18}{38}$, $p_{\text{black}}=\dfrac{18}{38}$, and $p_{\text{green}}=\dfrac{2}{38}$. All of these proportions combined equal 1. $H_0: p_{\text{red}}=\dfrac{18}{38}$, $p_{\text{black}}=\dfrac{18}{38}$, and $p_{\text{green}}=\dfrac{2}{38}$. $H_a$: At least one $p_i$ is not as specified in the null. In order to conduct a chi-square goodness of fit test all expected values must be at least 5. For both red and black: Expected count $=100(\dfrac{18}{38})=47.368$. For green: Expected count $=100(\dfrac{2}{38})=5.263$. All expected counts are at least 5 so we can conduct a chi-square goodness of fit test. 2. **Compute the test statistic.** $\chi^2=\sum \dfrac{(Observed-Expected)^2}{Expected}$ In the first step we computed the expected values for red and black to be 47.368 and for green to be 5.263. $\chi^2= \dfrac{(44-47.368)^2}{47.368}+\dfrac{(49-47.368)^2}{47.368}+\dfrac{(7-5.263)^2}{5.263}$ $\chi^2=0.239+0.056+0.573=0.868$ 3. **Determine the p-value.** Our sampling distribution will be a chi-square distribution. $df=k-1=3-1=2$ We can find the p-value by constructing a chi-square distribution with 2 degrees of freedom to find the area to the right of $\chi^2=0.868$ ![](assets/ChiSqDist2dfRight686.png){#fig-ChiSqDist2dfRight686 fig-alt="Chi-squared distribution plot made using Minitab Express; degrees of freedom equal 2; area to the right of chi-squared value of 0.868 is shaded with a proportion of 0.647912" .mx-auto .d-block width="60%" .lightbox} The p-value is 0.647912. 4. **Make a decision.** $p>0.05$ therefore we should fail to reject the null hypothesis. 5. **State a "real world" conclusion.** There is not enough evidence that this roulette wheel is unfair. ::: ::: ### ![ ](assets/minitab.png){.icon-image} Minitab: Goodness-of-Fit Test {{< include _minitab/conduct-a-chi-square-goodness-of-fit-test-with-summarized-data.qmd >}} ::: {#exm-cards-goodness-of-fit-in-minitab .example-minitab} #### Minitab: Cards (Equal Proportions) **Research Question**: When randomly selecting a card from a deck with replacement, are we equally likely to select a heart, diamond, spade, and club? I randomly selected a card from a standard deck 40 times with replacement. I pulled 13 Hearts (♥), 8 Diamonds (♦), 8 Spades (♠), and 11 Clubs (♣). To perform a chi-square goodness-of-fit test in Minitab using summarized data we first need to enter the data into the worksheet. Below you can see that we have one column with the names of each group and one column with the observed counts for each group. ```{=html} <table align="center" class="table w-auto mx-auto table-sm table-responsive bordered" data-quarto-disable-processing="false"> <thead class="bg-light text-black"> <tr> <th class="text-center" style="width: 5%;"></th> <th class="text-center" style="width: 5%;">C1</th> <th class="text-center" style="width: 5%;">C2</th> </tr> </thead> <tbody> <tr class="text-end"> <th></th> <th>Suit</th> <th>Count</th> </tr> <tr class="text-end"> <td>1</td> <th>Hearts</th> <td>13</td> </tr> <tr class="text-end"> <td>2</td> <th>Diamonds</th> <td>8</td> </tr> <tr class="text-end"> <td>3</td> <th>Spades</th> <td>8</td> </tr> <tr class="text-end"> <td>4</td> <th>Clubs</th> <td>11</td> </tr> </tbody> </table> ``` 1. After entering the data, select **Stat > Tables > Chi-Square Goodness of Fit Test (One Variable)**. 2. Insert *Count* into the **Observed Counts** box. 3. Insert *Suit* into the **Category names (optional)** box. 4. Choose **OK**. This should result in the following output: ::: {.minitab_output} #### Chi-Square Goodness-of-Fit Test: Count {.unnumbered .unlisted} #### Observed and Expected Counts {.unnumbered .unlisted} | Category | Observed | TestProportion | Expected | Contributionto Chi-Sq | | --- |--- |--- |--- |--- | | Hearts | 13 | 0.25 | 10 | 0.9 | | Diamonds | 8 | 0.25 | 10 | 0.4 | | Spades | 8 | 0.25 | 10 | 0.4 | | Clubs | 11 | 0.25 | 10 | 0.1 | : {.w-auto .table-sm .row-header} #### Chi-Square Test {.unnumbered .unlisted} | N | DF | Chi-Sq | P-Value | | --- |--- |--- |--- | | 40 | 3 | 1.8 | 0.615 | : {.w-auto .table-sm .row-header} ::: All expected values are at least 5 so we can use the chi-square distribution to approximate the sampling distribution. Our results are $\chi^2 (3) = 1.8$. $p = 0.615$. Because our p-value is greater than the standard alpha level of 0.05, we fail to reject the null hypothesis. There is not enough evidence to conclude that the proportions are different in the population. ::: ::: {.callout-caution appearance="minimal"} #### Note! The example above tested equal population proportions. Minitab also has the ability to conduct a chi-square goodness-of-fit test when the hypothesized population proportions are not all equal. To do this, you can choose to test specified proportions or to use proportions based on historical counts. ::: ::: {#exm-tulips .example-minitab} #### Tulips (Summarized Data, Equal Proportions) A company selling tulip bulbs claims they have equal proportions of white, pink, and purple bulbs and that they fill customer orders by randomly selecting bulbs from the population of all of their bulbs. You ordered 30 bulbs and received 16 white, 8 pink, and 6 purple. Is there convincing evidence the bulbs you received were not randomly selected from a population with an equal proportion of each color? Use Minitab to conduct a hypothesis test to address this research question. We'll go through each of the steps in the hypotheses test: ::: {.steps} 1. **Check assumptions and write hypotheses.** $H_0\colon p_{white}=p_{pink}=p_{purple}=\dfrac{1}{3}$ $H_a\colon$ at least one $p_i$ is not $\dfrac{1}{3}$ We can use the null hypothesis to check the assumption that all expected counts are at least 5. $Expected\;count=n (p_i)$ All $p_i$ are $\frac{1}{3}$. $30(\frac{1}{3})=10$, thus this assumption is met and we can approximate the sampling distribution using the chi-square distribution. 2. **Compute the test statistic.** Let's use Minitab to calculate this. First, enter the summarized data into a Minitab Worksheet. ```{=html} <table align="center" class="table w-auto mx-auto table-sm table-responsive bordered" data-quarto-disable-processing="false"> <thead class="bg-light text-black"> <tr> <th class="text-center" style="width: 5%;"></th> <th class="text-center" style="width: 5%;">C1</th> <th class="text-center" style="width: 5%;">C2</th> </tr> </thead> <tbody> <tr class="text-end"> <th></th> <th>Color</th> <th>Count</th> </tr> <tr class="text-end"> <td>1</td> <th>White</th> <td>16</td> </tr> <tr class="text-end"> <td>2</td> <th>Pink</th> <td>8</td> </tr> <tr class="text-end"> <td>3</td> <th>Purple</th> <td>6</td> </tr> </tbody> </table> ``` 1. After entering the data, select **Stat > Tables > Chi-Square Goodness of Fit Test (One Variable)**. 2. Insert *Count* into the **Observed Counts** box. 3. Insert *Color* into the **Category names (optional)** box. 4. Choose **OK**. This should result in the following output: ::: {.minitab_output} #### Chi-Square Goodness-of-Fit Test: Count {.unnumbered .unlisted} ##### Observed and Expected Counts {.unnumbered .unlisted} | Category | Observed | Test Proportion | Expected | Contribution to Chi-Sq | | --- |--- |--- |--- |--- | | White | 16 | 0.333333 | 10 | 3.6 | | Pink | 8 | 0.333333 | 10 | 0.4 | | Purple | 6 | 0.333333 | 10 | 1.6 | : {.w-auto .table-sm .row-header} ##### Chi-Square Test {.unnumbered .unlisted} | N | DF | Chi-Sq | P-Value | | --- |--- |--- |--- | | 30 | 2 | 5.6 | 0.061 | : {.w-auto .table-sm .row-header} The test statistic is a Chi-Square of 5.6. ::: 3. **Determine the p-value.** The p-value from the output is 0.061. 4. **Make a decision.** $p>0.05$ therefore we fail to reject the null hypothesis. 5. **State a "real world" conclusion.** There is not enough evidence that the tulip bulbs were not randomly selected from a population with equal proportions of white, pink and purple. ::: ::: ::: {#exm-roulette} #### Roulette (Summarized Data, Different Proportions) An American roulette wheel contains 38 slots: 18 red, 18 black, and 2 green. A casino has purchased a new wheel and they want to know if there is convincing evidence that the wheel is unfair. They spin the wheel 100 times and it lands on red 44 times, black 49 times, and green 7 times. Use Minitab to conduct a hypothesis test to address this question. We'll go through each of the steps in the hypotheses test: ::: {.steps} 1. **Check assumptions and write hypotheses.** If the wheel is 'fair' then the probability of red and black are both 18/38 and the probability of green is 2/38. $H_0\colon p_{red}=\dfrac{18}{38}, p_{black}=\dfrac{18}{38}, p_{green}=\dfrac{2}{38}$ $H_a\colon$ at least one $p_i$ is not as specified in the null We can use the null hypothesis to check the assumption that all expected counts are at least 5. $\text{Expected count}=n (p_i)$ With n = 100 we meet the assumptions needed to use Chi-square. 2. **Compute the test statistic.** Let's use Minitab to calculate this. First, enter the summarized data into a Minitab Worksheet. ```{=html} <table align="center" class="table w-auto mx-auto table-sm table-responsive bordered" data-quarto-disable-processing="false"> <thead class="bg-light text-black"> <tr> <th class="text-center" style="width: 5%;"></th> <th class="text-center" style="width: 5%;">C1</th> <th class="text-center" style="width: 5%;">C2</th> <th class="text-center" style="width: 5%;">C3</th> </tr> </thead> <tbody> <tr class="text-end"> <th></th> <th>Like Tea</th> <th>Like Coffee—Yes</th> <th>Like Coffee—No</th> </tr> <tr class="text-end"> <td>1</td> <th>Yes</th> <td>30</td> <td>25</td> </tr> <tr class="text-end"> <td>2</td> <th>No</th> <td>10</td> <td>35</td> </tr> </tbody> </table> ``` 1. After entering the data, select **Stat > Tables > Chi-Square Goodness of Fit Test (One Variable)**. 2. Insert *Count* into the **Observed Counts** box. 3. Insert *Color* into the **Category names (optional)** box. 4. For **Test**, select **Input constants**. 5. Select **Proportions specified by historical counts** (this is what we would expect if the null was true). 6. Enter 18/38 for *Black*, 2/38 for *Green*, and 18/38 for *Red*. 7. Choose **OK**. This should result in the following output: ::: {.minitab_output} #### Chi-Square Goodness-of-Fit Test: Count {.unnumbered .unlisted} ##### Observed and Expected Counts {.unnumbered .unlisted} | Category | Observed | Historical Counts | Test Proportion | Expected | Contribution to Chi-Sq | | --- |--- |--- |--- |--- |--- | | Red | 44 | 18 | 0.473684 | 47.3684 | 0.239532 | | Black | 49 | 18 | 0.473684 | 47.3684 | 0.056199 | | Green | 7 | 2 | 0.052632 | 5.2632 | 0.573158 | : {.w-auto .table-sm .row-header} ##### Chi-Square Test {.unnumbered .unlisted} | N | DF | Chi-Sq | P-Value | | --- |--- |--- |--- | | 100 | 2 | 0.868889 | 0.648 | : {.w-auto .table-sm .row-header} The test statistic is a Chi-Square of 0.87. ::: 3. **Determine the p-value.** The p-value from the output is 0.648. 4. **Make a decision.** $p>0.05$ therefore we fail to reject the null hypothesis. 5. **State a "real world" conclusion.** There is not enough evidence to state that this roulette wheel is unfair. ::: ::: ## Chi-Square Test of Independence The chi-square ($\chi^2$) test of independence is used to test for a relationship between two categorical variables. Recall that if two categorical variables are independent, then $P(A) = P(A \mid B)$. The chi-square test of independence uses this fact to compute expected values for the cells in a two-way contingency table under the assumption that the two variables are independent (i.e., the null hypothesis is true). Even if two variables are independent in the population, samples will vary due to random sampling variation. The chi-square test is used to determine if there is convincing evidence that the two variables are not independent in the population using the same hypothesis testing logic that we used with one mean, one proportion, etc. Again, we will be using the five-step hypothesis testing procedure: ::: {.steps} 1. **Check assumptions and write hypotheses.** The assumptions are that the sample is randomly drawn from the population and that all expected values are at least 5 (we will see what expected values are later). Our hypotheses are: $H_0:$ There is not a relationship between the two variables in the population (they are independent) $H_a:$ There is a relationship between the two variables in the population (they are dependent) Note: When you're writing the hypotheses for a given scenario, use the names of the variables, not the generic "two variables." 2. **Calculate an appropriate test statistic.** ::: {.callout-note appearance="minimal"} [**Chi-Square Test Statistic**]{.fs-4} $$ \chi^2=\sum \dfrac{(\text{Observed}-\text{Expected})^2}{\text{Expected}} $$ ::: ::: {.callout-note appearance="minimal"} [**Expected Cell Value**]{.fs-4} $$ E=\dfrac{\text{row total} \times \text{column total}}{n} $$ ::: 3. **Determine the p-value.** The p-value can be found using Minitab. Look up the area to the right of your chi-square test statistic on a chi-square distribution with the correct degrees of freedom. Chi-square tests are always right-tailed tests. ::: {.callout-note appearance="minimal"} [**Degrees of Freedom: Chi-Square Test of Independence**]{.fs-4} $df=(\text{number of rows}-1)(\text{number of columns}-1)$ ::: 4. **Make a decision.** If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis. 5. **State a "real world" conclusion.** Write a conclusion in terms of the original research question. ::: ::: {#exm-gender-and-online-learning} ### Gender and Online Learning A sample of 314 Penn State students was asked if they have ever taken an online course. Their genders were also recorded. The contingency table below was constructed. Use a chi-square test of independence to determine if there is a relationship between gender and whether or not someone has taken an online course. | | Yes | No | ----- | -------------------------------- | ---- | | Men | 43 | 63 | | Women | 95 | 113 | : Have you taken an online course? {.w-auto .table-sm .mx-auto .row-header} #### Solution {.unnumbered .unlisted} ::: {.steps} 1. **Check assumptions and write hypotheses.** $H_0:$ There is not a relationship between gender and whether or not someone has taken an online course (they are independent) $H_a:$ There is a relationship between gender and whether or not someone has taken an online course (they are dependent) Looking ahead to our calculations of the expected values, we can see that all expected values are at least 5. This means that the sampling distribution can be approximated using the $\chi^2$ distribution. 2. **Calculate the test statistic.** In order to compute the chi-square test statistic we must know the observed and expected values for each cell. We are given the observed values in the table above. We must compute the expected values. The table below includes the row and column totals. | | Yes | No | Total | | --- |--- |--- | ---| | Men | 43 | 63 | 106 | | Women | 95 | 113 | 208 | | Total| 138 | 176 | 314 | : Have you taken an online course? {.w-auto .table-sm .mx-auto .row-header} $E=\dfrac{row\;total \times column\;total}{n}$ \ $E_{\text{Men, Yes}}=\dfrac{106\times138}{314}=46.586$ \ $E_{\text{Men, No}}=\dfrac{106\times176}{314}=59.414$ \ $E_{\text{Women, Yes}}=\dfrac{208\times138}{314}=91.414$ \ $E_{\text{Women, No}}=\dfrac{208 \times 176}{314}=116.586$ Note that all expected values are at least 5, thus this assumption of the $\chi^2$ test of independence has been met. Observed and expected counts are often presented together in a contingency table. In the table below, expected values are presented in parentheses. | | Yes | No | Total | --- |--- |--- | ---| | Men | 43 (46.586) | 63 (59.414) | 106 | | Women | 95 (91.414) | 113 (116.586) | 208 | | Total | 138 | 176 | 314 | : Have you taken an online course? {.w-auto .table-sm .mx-auto .row-header} $$ \chi^2=\sum \dfrac{(O-E)^2}{E} $$ $$ \begin{align*} \chi^2&=\dfrac{(43-46.586)^2}{46.586}+\dfrac{(63-59.414)^2}{59.414}+\dfrac{(95-91.414)^2}{91.414}+\dfrac{(113-116.586)^2}{116.586} \\ &=0.276+0.216+0.141+0.110 \\ &=0.743 \end{align*} $$ The chi-square test statistic is 0.743. $$ df=(\text{number of rows}-1)(\text{number of columns}-1)=(2-1)(2-1)=1 $$ 3. **Determine the p-value.** We can determine the p-value by constructing a chi-square distribution plot with 1 degree of freedom and finding the area to the right of 0.743. ![](assets/ChiSq1ForStat743.png){#fig-ChiSq1ForStat743 fig-alt="Distribution Plot - Chi-Square, DF=1" .mx-auto .d-block width="60%" .lightbox} $p = 0.388702$ 4. **Make a decision.** $p>\alpha$, therefore we fail to reject the null hypothesis. 5. **State a "real world" conclusion.** There is not enough evidence to conclude that gender and whether or not an individual has completed an online course are related. Note that we cannot say for sure that these two categorical variables are independent, we can only say that we do not have enough evidence that they are dependent. ::: ::: ### ![ ](assets/minitab.png){.icon-image} Minitab: Test of Independence #### Raw vs Summarized Data {.unnumbered .unlisted} If you have a data file with the responses for individual cases then you have "raw data" and can follow the directions below. If you have a table filled with data, then you have "summarized data." There is an example of conducting a chi-square test of independence using summarized data in a later section. After data entry the procedure is the same for both data entry methods. #### ![ ](assets/minitab.png){.icon-image} Minitab: Chi-square Test Using Raw Data {{< include _minitab/conduct-a-chi-square-test-of-association.qmd >}} ::: {#exm-seat-location-and-cheating-in-minitab .minitab-example} #### Minitab: Seat Location and Cheating **Research question**: Is there a relationship between where a student sits in class and whether they have ever cheated? - **Null hypothesis**: Seat location and cheating are not related in the population. - **Alternative hypothesis**: Seat location and cheating are related in the population. To perform a chi-square test of independence in Minitab using raw data: 1. Open Minitab file: ([class_survey.mpx](data_files/class_survey.mpx){download="" target="_blank"}) 2. Select **Stat > Tables > Chi-Square Test for Association**. 3. Select **Raw data (categorical variables)** from the dropdown menu. 4. Insert the variable *Seating* into the **Rows** box. 5. Insert the variable *Ever\_Cheat* into the **Columns** box. 6. Select **Statistics** and check the boxes **Chi-square test for association** and **Expected cell counts**. 7. Choose **OK** and **OK**. This should result in the following output: ::: {.minitab_output} #### Rows: Seating Columns: Ever_Cheat {.unnumbered .unlisted} | | No | Yes | All | | ------ | ------------- | ------ | --- | | Back | 24 <br>24.21 | 8 <br>7.79 | 32 | | Front | 38 <br>34.81 | 8 <br>11.19 | 46 | | Middle | 109 <br>111.98 | 39 <br>36.02 | 148 | | All | 1714 | 55 | 226 | : {.w-auto .table-sm .row-header} *Cell contents* : *Count* *Expected count* #### Chi-Square Test {.unnumbered .unlisted} | | Chi-Square | DF | P-Value | | --- |--- |--- |--- | | Pearson | 1.539 | 2 | 0.463 | | Likelihood Ratio | 1.626 | 2 | 0.443 | : {.w-auto .table-sm .row-header} ::: #### Interpret {.unnumbered .unlisted} All expected values are at least 5 so we can use the Pearson chi-square test statistic. Our results are $\chi^2 (2) = 1.539$. $p = 0.463$. Because our $p$ value is greater than the standard alpha level of 0.05, we fail to reject the null hypothesis. There is not enough evidence of a relationship in the population between seat location and whether a student has cheated. ::: ::: {#exm-dog-cat-ownership .minitab-example} #### Dog & Cat Ownership (Raw Data) Is there a relationship between dog and cat ownership in the population of all World Campus STAT 200 students? Let's conduct a hypothesis test using the dataset: ([fall2016stdata.mpx](data_files/fall2016stdata.mpx){download="" target="_blank"}) ::: {.steps} 1. **Check any necessary assumptions and write null and alternative hypotheses.** $H_0:$ There is not a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students $H_a:$ There is a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students Assumption: All expected counts are at least 5. The expected counts here are 176.02, 75.98, 189.98, and 82.02, so this assumption has been met. 2. **Calculate an appropriate test statistic.** Let's use Minitab to calculate the test statistic and p-value. 1. After entering the data, select **Stat > Tables > Cross Tabulation and Chi-Square**. 2. Enter *Dog* into the **Rows** box. 3. Enter *Cat* in the **Columns** box. 4. Select the **Chi-Square** button, and in the new window, check the box for the **Chi-square test** and **Expected cell counts**. 5. Choose **OK** and **OK**. ::: {.minitab_output} #### Rows: Dog Columns: Cat {.unnumbered .unlisted} | | No | Yes | All | | ------- | --------------- | ------- | --- | | No | 183 <br> 176.02 | 69 <br> 75.98 | 252 | | Yes | 183 <br> 189.98 | 89 <br> 82.02 | 272 | | Missing | 1 | 0 | | | All | 366 | 158 | 524 | : {.w-auto .table-sm .row-header} *Cell contents:* : *Count* *Expected count* ##### Chi-Square Test {.unnumbered .unlisted} | | Chi-Square | DF | P-Value | | --- |--- |--- |--- | | Pearson | 1.771 | 1 | 0.183 | | Likelihood Ratio | 1.775 | 1 | 0.183 | : {.w-auto .table-sm .row-header} ::: Since the assumption was met in step 1, we can use the Pearson chi-square test statistic. $\text{Pearson's}\chi^2 = 1.771$ 3. **Determine a p-value associated with the test statistic.** $p = 0.183$ 4. **Decide between the null and alternative hypotheses.** Our p-value is greater than the standard 0.05 alpha level, so we fail to reject the null hypothesis. 5. **State a "real world" conclusion.** There is not enough evidence of a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students. ::: ::: ::: {#exm-coffee-and-tea-preference .minitab-example} #### Coffee and Tea Preference (Summarized Data) Is there a relationship between liking tea and liking coffee? The following table shows data collected from a random sample of 100 adults. Each were asked if they liked coffee (yes or no) and if they liked tea (yes or no). | Tea preference | Likes coffee | Does not like coffee | |:-------------------|-------------:|---------------------:| | Likes tea | 30 | 25 | | Does not like tea | 10 | 35 | : Coffee and tea preferences {.w-auto .table-sm .mx-auto .row-header} Let's use the five-step hypothesis testing procedure to address this research question. ::: {.steps} 1. **Check any necessary assumptions and write null and alternative hypotheses.** $H_0:$ Liking coffee and liking tea are not related (i.e., independent) in the population $H_a:$ Liking coffee and liking tea are related (i.e., dependent) in the population Assumption: All expected counts are at least 5. 2. **Calculate an appropriate test statistic.** Let's use Minitab to calculate the test statistic and p-value. 1. Enter the table into a Minitab worksheet as shown below: ::: {.w-75 .mx-auto} ```{=html} <table align="center" class="table w-auto mx-auto table-sm table-responsive bordered" data-quarto-disable-processing="false"> <thead class="bg-light text-black"> <tr> <th class="text-center" style="width: 5%;"></th> <th class="text-center" style="width: 5%;">C1</th> <th class="text-center" style="width: 5%;">C2</th> <th class="text-center" style="width: 5%;">C3</th> </tr></thead> <tbody> <tr class="text-end"> <th></th> <th>Likes Tea</th> <th>Likes Coffee-Yes</th> <th>Likes Coffee-No</th> </tr> <tr class="text-end"> <td>1</td> <th>Yes</th> <td>30</td> <td>25</td> </tr> <tr class="text-end"> <td>2</td> <th>No</th> <td>10</td> <td>35</td> </tr> </tbody></table> ``` ::: 2. Select **Stat > Tables > Cross Tabulation and Chi-Square**. 3. Select **Summarized data in a two-way table** from the dropdown. 4. Enter the columns *Likes Coffee-Yes* and *Likes Coffee-No* in the Columns containing the table box. 5. For the row labels, enter *Likes Tea* (leave the column labels blank). 6. Select the **Chi-Square** button and check the boxes for **Chi-square test** and **Expected cell counts**. 7. Choose **OK** and **OK**. **Output** ::: {.minitab_output} #### Rows: Likes Tea Columns: Worksheet columns {.unnumbered .unlisted} | | No | Yes | All | | --- | --------- | --------- | --- | | Yes | 30 <br>22 | 25 <br>33 | 55 | | No | 10 <br>18 | 35 <br>27 | 45 | | All | 40 | 60 | 100 | : {.w-auto .table-sm .row-header} *Cell contents:* : *Count* *Expected count* ##### Chi-Square Test {.unnumbered .unlisted} | | Chi-Square | DF | P-Value | | --- |--- |--- |--- | | Pearson | 10.774 | 1 | 0.001 | | Likelihood Ratio | 11.138 | 1 | 0.001 | : {.w-auto .table-sm .row-header} ::: Since the assumption was met in step 1, we can use the Pearson chi-square test statistic. $\text{Pearson's }\chi^2 = 10.774$ 1. **Determine a p-value associated with the test statistic.** $p = 0.001$ 2. **Decide between the null and alternative hypotheses.** Our p-value is less than the standard 0.05 alpha level, so we reject the null hypothesis. 3. **State a "real world" conclusion.** There is convincing evidence of a relationship between liking coffee and liking tea in the population. ::: ::: ### Relative Risk A chi-square test of independence will give you information concerning whether or not a relationship between two categorical variables in the population is likely. As was the case with the single sample and two sample hypothesis tests that you learned earlier this semester, with a large sample size statistical power is high and the probability of rejecting the null hypothesis is high, even if the relationship is relatively weak. In addition to examining statistical significance by looking at the p-value, we can also examine practical significance by computing the **relative risk**. In [Lesson 2](Lesson02.qmd) you learned that risk is often used to describe the probability of an event occurring. Risk can also be used to compare the probabilities in two different groups. First, we'll review risk, then you'll be introduced to the concept of relative risk. ::: {#def-risk} #### Risk The **risk** of an outcome can be expressed as a fraction or as the percent of a group that experiences the outcome. $$ \text{Risk}=\dfrac{\text{n with outcome}}{\text{total n}} $$ ::: ::: {#exm-examples-of-risk} #### Examples of Risk **Asthma**: 60 out of 1000 teens have asthma. The risk is $\frac{60}{1000}=.06$. This means that 6% of all teens experience asthma. **Flu**: 45 out of 100 children get the flu each year. The risk is $\frac{45}{100}=.45$ or 45% ::: ::: {#def-relativerisk} #### Relative Risk Relative risk compares the risk of a particular outcome in two different groups. $$ \text{Relative Risk}=\dfrac{\text{Risk in Group 1}}{\text{Risk in Group 2}} $$ ::: Thus, relative risk gives the risk for group 1 as a multiple of the risk for group 2. ::: {#exm-example-of-relative-risk} #### Relative Risk for Flu in Children and Adults Suppose that the risk of a child getting the flu this year is .45 and the risk of an adult getting the flu this year is .10. What is the relative risk of children compared to adults? $$ \text{Relative risk}=\dfrac{.45}{.10}=4.5 $$ Children are 4.5 times more likely than adults to get the flu this year. ::: ::: {.caution-block .rounded-3 .p-3} #### []{.bi .bi-exclamation-triangle-fill .fs-4 .text-warning} Cautions {.unnumbered .unlisted .mt-0} Watch out for relative risk statistics where no baseline information is given about the actual risk. For instance, it doesn't mean much to say that beer drinkers have twice the risk of stomach cancer as non-drinkers unless we know the actual risks. The risk of stomach cancer might actually be very low, even for beer drinkers. For example, 2 in a million is twice the size of 1 in a million but is would still be a very low risk. This is known as the **baseline** with which other risks are compared. ::: ## Summary In this lesson you added two new statistical procedures to your repertoire: the chi-square goodness-of-fit test for one categorical variable and the chi-square test of independence for two variables. In the remaining weeks we will continue to learn more procedures.